Equation:
I=T/α= rFt/α
Substitution:
I=(.0207)(.881)/.249
I=.0732 kg m^2
Method 2:
I(pulley) .00058 kg m^2
Thin Rods: length .338m
mass .074 kg
I(thin rod) .0085 kg m^2 x4=.0339 kg m^2
Movable Mass: mass .186 kg
distance from center .338 m
I(movable mass) .0212 x4=.085 kg m^2
I=I(pulley)+I(rod)+I(movable masses)
I=.00058+.0339+.085=.11948 kg m^2
Experiment 3:
Gravitational PE:
mass: .09 kg
vertical displacement: .536 m
Equation: PE=mph
PE=(.09)(9.8)(.536)
PE=.473
Translational KE:
(t1) 14.8s (t2)13.7s (t3)14.2s (t avg) 14.23s (v avg) .038 m/s
Vf=2xVavg=2(.038)=.075
Equation:
KE=1/2mv^2
KE=1/2(.090)(.075)^2
KE=.000255
Experiment 1:
Method 1:
To find the angular acceleration of the falling mass I used the equation:
α = 2θ/t^2
After substitution:
=2.8π/14.23^2
=.249 rad/s^2
Method 2:
Vertical Displacement: .536m
Equation:
a=2x/t^2
After Substitution:
a=2(.536)/14.23^2
a=.00529 m/s^2
Equation:
α=aT/r
α=.255 rad/s^2
Experiment 2:
Method 1:
Angular Acceleration of rotating apparatus: .249 rad/s^2
Linear Acceleration of falling mass: .256 m/s^2
Radius of Pulley: .0207m
Falling Mass: .09kg
Weight of falling mass: .882N
Equation:
Ft=W-ma
Substitution:
Ft=(.882)-.09(.00529)
Ft=.881N
Equation:
I=T/α= rFt/α
Substitution:
I=(.0207)(.881)/.249
I=.0732 kg m^2
Method 2:
I(pulley) .00058 kg m^2
Thin Rods: length .338m
mass .074 kg
I(thin rod) .0085 kg m^2 x4=.0339 kg m^2
Movable Mass: mass .186 kg
distance from center .338 m
I(movable mass) .0212 x4=.085 kg m^2
I=I(pulley)+I(rod)+I(movable masses)
I=.00058+.0339+.085=.11948 kg m^2
Experiment 3:
Gravitational PE:
mass: .09 kg
vertical displacement: .536 m
Equation: PE=mph
PE=(.09)(9.8)(.536)
PE=.473
Translational KE:
(t1) 14.8s (t2)13.7s (t3)14.2s (t avg) 14.23s (v avg) .038 m/s
Vf=2xVavg=2(.038)=.075
Equation:
KE=1/2mv^2
KE=1/2(.090)(.075)^2
KE=.000255
Radius of Pulley: .0207m
w=12.82 rad/s
I=.1986kg m^2
Equation: KE=1/2 IV^2
KE=1/2(.0723)(3.62)^2
KE=.4737J
KE(total)=.000255(trans)+.4737(rot)=.4737J
The values were very close with a small percent error.